What is the solution for (D^2+13D+13)x=12t^2+2t-cos(t) ?

The solution for (D^2+13D+13)x=12t^2+2t-cos(t) is

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(D^2+13D+13)x=12t^2+2t-cos(t)

(D^{2} + 13 D + 13) x = 12 t^{2} + 2 t - cos (t)

x = c_{1} e^{\frac{( - 13 + 3 13 ) t}{2}} + c_{2} e^{\frac{( - 13 - 3 13 ) t}{2}} + \frac{12 t ^{2} - 22 t}{13} + \frac{262}{169} - \frac{13}{313} sin (t) - \frac{12}{313} cos (t)

Solution steps

(D^{2} + 13 D + 13) x = 12 t^{2} + 2 t - cos (t)

Solve linear ODE:

x = c_{1} e^{\frac{( - 13 + 3 13 ) t}{2}} + c_{2} e^{\frac{( - 13 - 3 13 ) t}{2}} + \frac{12 t ^{2} - 22 t}{13} + \frac{262}{169} - \frac{13}{313} sin (t) - \frac{12}{313} cos (t)

x = c_{1} e^{\frac{( - 13 + 3 13 ) t}{2}} + c_{2} e^{\frac{( - 13 - 3 13 ) t}{2}} + \frac{12 t ^{2} - 22 t}{13} + \frac{262}{169} - \frac{13}{313} sin (t) - \frac{12}{313} cos (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = x e^{\frac{x}{2}}

y^{^{''}} + 9 y = e^{2 x} + x^{2} sin (x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 4 + x^{- 1} e^{x}

y^{^{''}} + y = x e^{x} cos (2 x)

What is the solution for (D^2+13D+13)x=12t^2+2t-cos(t) ?
The solution for (D^2+13D+13)x=12t^2+2t-cos(t) is